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Question

Ten persons, amongst whom are $$A,B$$ and $$C$$ are to speak at a function. The number of ways in which it can be done if $$A$$ wants to speak before $$B$$, and $$B$$ wants to speak before $$C$$ is 


A
10!6
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B
8!
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C
8!3
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D
3!
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Solution

The correct option is D $$\displaystyle\frac{10!}{6}$$
Places for A, B and C can be chosen in $$^{10}C_3$$ ways
Remaining 7 can be selected in $$7!$$ ways
So total $$= ^{10}C_3\times 7!$$
$$=\displaystyle \frac{10!.7!}{3!.7!}$$
$$=\displaystyle \frac{10!}{6}$$
Hence, option 'A' is correct.

Maths

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